If the internal energy of an ideal gas decreases by the same amount as the work done by the system, then the

A. Process must be isobaric

B. Temperature must decrease

C. Process must be adiabatic

D. Both B and C

Answer: Option D

Solution(By Examveda Team)

By first law of thermodynamics we can say $$\delta Q = dU + \delta W$$
$$ \Rightarrow $$  for the internal energy and work to be same in magnitude we can say that heat transfer has to be zero $$\delta Q = 0$$
So the process is adiabatic and given the system is ideal gas and its internal energy is decreasing so, we can say that temperature decreases since the internal energy of a substance is solely dependent of temperature.